Products after sorting Given two sets $\{a_1, \dots, a_N\}$ and $\{ b_1, \dots, b_N \}$ of positive numbers $0 < a_i, b_i < 1$, $i=1, \dots, N$, is it true that $\sum_i \min(a_i, b_i) \cdot \sum_i \max(a_i, b_i) \le \sum_i a_i \cdot \sum_i b_i$?
 A: A cumbersome way to prove this is by induction, and it seems it holds not only for probabilities, but also for all integers. It's apparent that this holds for $N=1$. Now, assume it holds for $N=n-1$, and let's query for $N=n$. For taking advantage of the result for $N=n-1$, we arrange the terms as below: 
$$\begin{align}&\left(\sum_{i=1}^{n-1}\min(a_i,b_i)+\min(a_n,b_n)\right)\left(\sum_{i=1}^{n-1}\max(a_i,b_i)+\max(a_n,b_n)\right) \\ &\overbrace{\leq}^?\left(\sum_{i=1}^{n-1}a_i+a_n\right)\left(\sum_{i=1}^{n-1}b_i+b_n\right)\end{align}$$
We expand the terms and since the inequality holds for $n-1$, the part that needs proving is (assuming $a_n\leq b_n$ without loss of generality):
$$a_n\sum_{i=1}^{n-1}\max(a_i,b_i)+b_n\sum_{i=1}^{n-1}\min(a_i,b_i)\overbrace{\leq}^? a_n\sum_{i=1}^{n-1}b_i+b_n\sum_{i=1}^{n-1}a_i$$
$$a_n\sum_{i=1}^{n-1}(\max(a_i,b_i)-b_i)\overbrace{\leq}^? b_n\sum_{i=1}^{n-1}(a_i-\min(a_i,b_i))$$
Now, consider the LHS and RHS sums. When $a_i\leq b_i$, the summands are $0$. Then, we're only left with positive terms in indices, $I$, where $a_i>b_i$, and the sum reduces to:
$$a_n\sum_{I}(a_i-b_i)\overbrace{\leq}^? b_n\sum_I (a_i-b_i)\rightarrow a_n\overbrace{\leq}^? b_n$$
which is our initial assumption. Logically, one can reach a true conclusion from a wrong statement, however these operations can be replicated backwards, which makes it a $\iff$ statement. So, the inequality is true. Note that, I just assumed $a_n\leq b_n$ for notational simplicity. The expression in the end would look like $\min(a_n,b_n)\leq \max(a_n,b_n)$ if I didn't assume $a_n<b_n$, and the lines would look more populated. 
